Physics 313
Problem Set 2
Due Friday September 22, 2006 by the end of class
1. Schroeder Problem 1.36
A. Use
V
γ
P
=constant to get (1
L
)
γ
(1
atm
) =
V
γ
(7
atm
), so
V
= (1
/
7)
(1
/γ
)
.
Here
γ
= 1
.
4 for air, so
V
= 0
.
249 L.
B. To get the work, do
W
=
R
P
(
V
)
dV
. Now
V
γ
P
=constant = 1 to start
with (in units of literxs and atmospheres), so P =
V

γ
.
W
=

Z
V
f
V
i
dV V

γ
=
1
γ

1
V

γ
+1
f

V

γ
+1
i
=
1
0
.
4
(
0
.
249

0
.
4

1

0
.
4
)
= 1
.
86
So
W
= 1
.
86 L
·
atm, which is 188 J.
C. I assume the initial temperature is 300 K. The volume decreased from 1 L
to 0.249 L, but the pressure increased by a factor of 7. Using the ideal gas law,
the final temperature should scale with P and V, and so
T
= (300
K
)(0
.
249)(7) =
523 K.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2. Schroeder Problem 1.48
The incident power is 1000 W/m
2
, but 90% is reflected, so the absorbed
power is 100 W/m
2
.
The snow pack is 2 m thick.
Imagine a 1 m
2
patch of
the snow.
We’re told it’s 50% air, so the total volume of ice to be melted is
(2)(0
.
5) = 1 m
2
= 10
6
cm
2
.
Ice has a densityof 0.9, so this is 0
.
9
×
10
6
=
9
×
10
5
grams of ice to melt. Ice has a latent heat of 333 J/g, so melting all of
the ice would require (333)(9
×
10
5
) = 3
×
10
8
J. The ice pack would therefore
last for (3
×
10
8
J
)
/
100
W
= 3
×
10
6
seconds. In the spring (around equinox),
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 0000
 Work, total number, Schroeder, Schroeder Problem

Click to edit the document details